VCE Specialist 3/4 Question Thread!

[Will T]

Air resistance is an arbitrary thing, it is not a constant and changes between questions.

Eg in one question I might have a bullet, the other a couch. These will have different surface areas and thus different air resistances.
Mass of the particles can also be different.

I mean if you apply F = ma to example 22.
Choosing up to be positive (as they have done)
-mg+0.2v^2=ma

Implies -g + 0.2v^2/m = a

which is not -g + 0.2v^2 = a
(unless m = 1 kg, and it doesn't say the mass).
Do you see what I mean when I say I feel that's inconsistent to your thought of applying F = ma here?
I feel we can conclude that the Textbook is wrong.... I was just pointing out something I thought was inconsistent, thanks anyway.

[Stevensmay]

I am pretty sure there is information missing/misprint in this case.
I cannot see how they remove the m constant.

The same f=ma idea can be applied here.

[b^3]

I am pretty sure there is information missing/misprint in this case.
I cannot see how they remove the m constant.

The same f=ma idea can be applied here.

They can remove the constant if it's contained in the , as they are adding accelerations and not forces. It's badly worded, they should have said 'the acceleration/deceleration due to air resistance is given by/can be modeled be ".

The way I read it the first time was that we were giving the force due to air resistance, rather than the acceleration, normally it's more clear.

[simba]

Hey, I've just got a quick question about ellipses! Is the major axis the x axis diameter or is it the longest diameter? I always assumed it was the x axis, but after I did a practice exam I realised that assumption was most likely wrong :/

[Stevensmay]

Hey, I've just got a quick question about ellipses! Is the major axis the x axis diameter or is it the longest diameter? I always assumed it was the x axis, but after I did a practice exam I realised that assumption was most likely wrong :/

Major axis is the one that is greater in magnitude.

http://www.mathopenref.com/ellipseaxes.html
http://www.mathsisfun.com/geometry/ellipse.html
http://www.thefreedictionary.com/major+axis

[ahat]

a = semi-major axis (horizontal axis, i.e. parallel to x-axis)
b - semi-minor axis (vertical axis, i.e. parallel to y-axis)

I have it on my wall
From my notes at least

[jono88]

Solve the differential equation dy/dx=1+x/(1-x)^2 given y=0 when x=0
have no clue what to do do O.O

Mod Edit: Merged double post (Phy124)

[b^3]

Hint: Integrate both sides and try a substitution, , then you'll end up with multiple terms on the top and a single term in the denominator, which will allow you to split the fraction up into nice things to integrate. Then remember to find with the initial conditions at the end.

[jono88]

ahh i think i got it.

Is it loge(1-x) +2/1-x -2?

[brightsky]

another way: dy/dx = 1+x/(1-x)^2 = 1 + [(x-1)+1]/(x-1)^2 = 1 + 1/(x-1) + 1/(x-1)^2. the integration is trivial from this point onwards.

[jono88]

With complex numbers, general solutions have (2k+1) where k is subset of Z. Can someone explain why this is the case?

[Stevensmay]

With complex numbers, general solutions have (2k+1) where k is subset of Z. Can someone explain why this is the case?

Polar form. It is because there are an infinite number of ways of representing the same angle, so we use the set k in place of listing every number.





But seeing as it sucks to have to write an infinite number of answers, we can take a short cut. By defining the set of all real integers to be k, we can just write our solution as

Complex is just slightly different due to where the solutions lie on the unit circle.
My apologies if I have misinterpreted your question.

[barydos]

Two questions here:

- Find a polynomial with real coefficients and with roots ""

- By expanding show that

[brightsky]

1. recall the fundamental theory of algebra which states that if "blah" is a root of a polynomial with real coefficients, then the conjugate of "blah" is also a root. now you just go and look for the roots with i in them, and find the complex conjugate of that. then you multiply them together to form the polynomial. in this case the polynomial is (z+2)(z-1-sqrt(2)i)(z-1+sqrt(2))(z-3i)(z+3i).
2. notice that the expression = 1. so expanding you get: cos^4 (x) + 2 cos^2(x) sin^2(x) + sin^4 (x) = 1
rearranging:
cos^4(x) + sin^4(x)
= 1 - 2 cos^2(x) sin^2(x)
= 1 - 2 (1/2+1/2cos(2x))(1/2-1/2cos(2x))
= 1 - 2(1/4 - 1/4 cos^2(2x))  [recall DOPS or whatever people call it these days]
= 1 - 1/2 + 1/2 cos^2(2x)
= 1/2 + 1/2 cos^2(2x)
= 1/2 + 1/2 (1/2 + 1/2 cos(4x))
= 1/2 + 1/4 + 1/4 cos(4x)
= 3/4 + 1/4 cos(4x)
yay

[barydos]

1. recall the fundamental theory of algebra which states that if "blah" is a root of a polynomial with real coefficients, then the conjugate of "blah" is also a root. now you just go and look for the roots with i in them, and find the complex conjugate of that. then you multiply them together to form the polynomial. in this case the polynomial is (z+2)(z-1-sqrt(2)i)(z-1+sqrt(2))(z-3i)(z+3i).
2. notice that the expression = 1. so expanding you get: cos^4 (x) + 2 cos^2(x) sin^2(x) + sin^4 (x) = 1
rearranging:
cos^4(x) + sin^4(x)
= 1 - 2 cos^2(x) sin^2(x)
= 1 - 2 (1/2+1/2cos(2x))(1/2-1/2cos(2x))
= 1 - 2(1/4 - 1/4 cos^2(2x))  [recall DOPS or whatever people call it these days]
= 1 - 1/2 + 1/2 cos^2(2x)
= 1/2 + 1/2 cos^2(2x)
= 1/2 + 1/2 (1/2 + 1/2 cos(4x))
= 1/2 + 1/4 + 1/4 cos(4x)
= 3/4 + 1/4 cos(4x)
yay

Damnnn you're a genius, thanks so much

Edit:
Another question:

Spoiler

With what reasoning can I eliminate option E?